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Proof step 21 from proof eq (previous step)

a ' + a0 ' + x0 = ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, a0 < a, y ≠ [] ∧ y.head < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
a ' + a0 ' + x0 = ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y = [], a0 < a, y.head < a

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' = b ' + x ↔ a = b ∧ x = []; (ws)